I'm trying to use stringr or R base calls to conditionally add a white-space for instances in a large vector where there is a numeric value then a special character - in this case a $ sign without a space. str_pad doesn't appear to allow for a reference vectors.
For example, for:
$6.88$7.34
I'd like to add a whitespace after the last number and before the next dollar sign:
$6.88 $7.34
Thanks!
If there is only one instance, then use sub to capture digit and the $ separately and in the replacement add the space between the backreferences of the captured group
sub("([0-9])([$])", "\\1 \\2", v1)
#[1] "$6.88 $7.34"
Or with a regex lookaround
gsub("(?<=[0-9])(?=[$])", " ", v1, perl = TRUE)
data
v1 <- "$6.88$7.34"
This will work if you are working with a vectored string:
mystring<-as.vector('$6.88$7.34 $8.34$4.31')
gsub("(?<=\\d)\\$", " $", mystring, perl=T)
[1] "$6.88 $7.34 $8.34 $4.31"
This includes cases where there is already space as well.
Regarding the question asked in the comments:
mystring2<-as.vector('Regular_Distribution_Type† Income Only" "Distribution_Rate 5.34%" "Distribution_Amount $0.0295" "Distribution_Frequency Monthly')
gsub("(?<=[[:alpha:]])\\s(?=[[:alpha:]]+)", "_", mystring2, perl=T)
[1] "Regular_Distribution_Type<U+2020> Income_Only\" \"Distribution_Rate 5.34%\" \"Distribution_Amount $0.0295\" \"Distribution_Frequency_Monthly"
Note that the \ appears due to nested quotes in the vector, should not make a difference. Also <U+2020> appears due to encoding the special character.
Explanation of regex:
(?<=[[:alpha:]]) This first part is a positive look-behind created by ?<=, this basically looks behind anything we are trying to match to make sure what we define in the look behind is there. In this case we are looking for [[:alpha:]] which matches a alphabetic character.
We then check for a blank space with \s, in R we have to use a double escape so \\s, this is what we are trying to match.
Finally we use (?=[[:alpha:]]+), which is a positive look-ahead defined by ?= that checks to make sure our match is followed by another letter as explained above.
The logic is to find a blank space between letters, and match the space, which then is replaced by gsub, with a _
See all the regex here
Related
I want to add a space between two punctuation characters (+ and -).
I have this code:
s <- "-+"
str_replace(s, "([:punct:])([:punct:])", "\\1\\s\\2")
It does not work.
May I have some help?
There are several issues here:
[:punct:] pattern in an ICU regex flavor does not match math symbols (\p{S}), it only matches punctuation proper (\p{P}), if you still want to match all of them, combine the two classes, [\p{P}\p{S}]
"\\1\\s\\2" replacement contains a \s regex escape sequence, and these are not supported in the replacement patterns, you need to use a literal space
str_replace only replaces one, first occurrence, use str_replace_all to handle all matches
Even if you use all the above suggestions, it still won't work for strings like -+?/. You need to make the second part of the regex a zero-width assertion, a positive lookahead, in order not to consume the second punctuation.
So, you can use
library(stringr)
s <- "-+?="
str_replace_all(s, "([\\p{P}\\p{S}])(?=[\\p{P}\\p{S}])", "\\1 ")
str_replace_all(s, "(?<=[\\p{P}\\p{S}])(?=[\\p{P}\\p{S}])", " ")
gsub("(?<=[[:punct:]])(?=[[:punct:]])", " ", s, perl=TRUE)
See the R demo online, all three lines yield [1] "- + ? =" output.
Note that in PCRE regex flavor (used with gsub and per=TRUE) the POSIX character class must be put inside a bracket expression, hence the use of double brackets in [[:punct:]].
Also, (?<=[[:punct:]]) is a positive lookbehind that checks for the presence of its pattern immediately on the left, and since it is non-consuming there is no need of any backreference in the replacement.
I have column names in the following format:
col= c('UserLanguage','Q48','Q21...20','Q22...21',"Q22_4_TEXT...202")
I would like to get the column names without everything that is after ...
[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
I am not sure how to code it. I found this post here but I am not sure how to specify the pattern in my case.
You can use gsub.
gsub("\\...*","",col)
#[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
Or you can use stringr
library(stringr)
str_remove(col, "\\...*")
Since . matches any character, we need to "escape" (\) to specify exactly what we want to match in the regular expression (and not use the special behavior of the .). So, to match a period, we would need \.. However, the backslash (\) is used to escape special behavior (e.g., escape symbol in strings) in regexps. So, to create the regular expression, we need an additional backslash, \\. In this case, we want to match additional periods, so we can add those here, hence \\.... Then, * specifies that the previous expression (everything the three periods) may occur 0 or more times.
You could sub and capture the first word in each column:
col <- c("UserLanguage", "Q48", "Q21...20", "Q22...21", "Q22_4_TEXT...202")
sub("^(\\w+).*$", "\\1", col)
[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
The regex pattern used here says to match:
^ from the start of the input
(\w+) match AND capture the first word
.* then consume the rest
$ end of the input
Then, using sub we replace with \1 to retain just the first word.
Surprisingly I haven't found a satisfactory answer to this regex problem. I have the following vector:
row1
[1] "AA.8.BB.CCCC" "2017" "3.166.5" "3.080.2" "68" "162.6"
[7] "185.223.632.4" "500.332.1"
My end result should look like this:
row1
[1] "AA.8.BB.CCCC" "2017" "3,166.5" "3,080.2" "68" "162.6"
[7] "185,223,632.4" "500,332.1"
The last period in each of the numeric values is the decimal point and the other periods should be converted to commas. I want this done without affecting the value with letters ([1]). I tried the following:
gsub("[.]\\d{3}[.]", ",", row1)
This regex sort of works but doesn't quite do what I want. Additionally it removes the numbers, which is problematic. Is there a way to find the regex and then only remove the first character and not the entire matched values? If there is a better way of approaching this I welcome those responses as well.
You can use the following:
See code in use here
gsub("\\G\\d+\\K\\.(?=\\d+(?!$))",",",x,perl=T)
See regex in use here
Note: The regex at the URL above is changed to (?:\G|^) for display purposes (\G matches the start of the string \A, but not the start of the line).
\G\d+\K\.(?=\d+(?!$))
How it works:
\G asserts position either at the end of the previous match or at the start of the string
\d+\K\. matches a digit one or more times, then resets the match (previously consumed characters are no longer included in the final match), then match a dot . literally
(?=\d+(?!$)) positive lookahead ensuring what follows is one or more digits, but not followed by the end of the line
One option is to use a combination of a lookbehind and a lookahead to match only a dot when what is on the left is a digit and on the right are 3 digits followed by a dot.
You could add perl = TRUE using gsub.
In the replacement use a comma.
(?<=\d)[.](?=\d{3}[.])
Regex demo | R demo
Double escaped as noted by #r2evans
(?<=\\d)[.](?=\\d{3}[.])
I have the following string:
x = "marchTextIWantToDisplayWithSpacesmarch"
I would like to delete the 'march' portion at the beginning of the string and then add a space before each uppercase letter in the remainder to yield the following result:
"Text I Want To Display With Spacesmarch"
To insert whitepace, I used gsub("([a-z]?)([A-Z])", "\\1 \\2", x, perl= T) but I have no clue how to modify the pattern so that the first 'march' is excluded from the returned string. I'm trying to get better at this so any help would be greatly appreciated.
An option would be to capture the upper case letter as a group ((...)) and in the replacement create a space followed by the backreference (\\1) of the captured group
gsub("([A-Z])", " \\1", x)
#[1] "march Text I Want To Display With Spacesmarch"
If we need to remove the 'march'
sub("\\b[a-z]\\w+\\s+", "", gsub("([A-Z])", " \\1", x))
[#1] "Text I Want To Display With Spacesmarch"
data
x <- "marchTextIWantToDisplayWithSpacesmarch"
No, you can't achieve your replacement using single gsub because in one of your requirement, you want to remove all lowercase letters starting from the beginning, and your second requirement is to introduce a space before every capital letter except the first capital letter of the resultant string after removing all lowercase letters from the beginning of text.
Doing it in single gsub call would have been possible in cases where somehow we can re-use some of the existing characters to make the conditional replace which can't be the case here. So in first step, you can use ^[a-z]+ regex to get rid of all lowercase letters only from the beginning of string,
sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch")
leaving you with this,
[1] "TextIWantToDisplayWithSpacesmarch"
And next step you can use this (?<!^)(?=[A-Z]) regex to insert a space before every capital letter except the first one as you might not want an extra space before your sentence. But you can combine both and write them as this,
gsub('(?<!^)(?=[A-Z])', ' ', sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch"), perl=TRUE)
which will give you your desired string,
[1] "Text I Want To Display With Spacesmarch"
Edit:
Explanation of (?<!^)(?=[A-Z]) pattern
First, let's just take (?=[A-Z]) pattern,
See the pink markers in this demo
As you can see, in the demo, every capital letter is preceded by a pink mark which is the place where a space will get inserted. But we don't want space to be inserted before the very first letter as that is not needed. Hence we need a condition in regex, which will not select the first capital letter which appears at the start of string. And for that, we need to use a negative look behind (?<!^) which means that Do not select the position which is preceded by start of string and hence this (?<!^) helps in discarding the upper case letter that is preceded by just start of string.
See this demo where the pink marker is gone from the very first uppercase letter
Hope this clarifies how every other capital letter is selected but not the very first. Let me know if you have any queries further.
You may use a single regex call to gsub coupled with trimws to trim the resulting string:
trimws(gsub("^\\p{Ll}+|(?<=.)(?=\\p{Lu})", " ", x, perl=TRUE))
## => [1] "Text I Want To Display With Spacesmarch"
It also supports all Unicode lowercase (\p{Ll}) and uppercase (\p{Lu}) letters.
See the R demo online and the regex demo.
Details
^\\p{Ll}+ - 1 or more lowercase letters at the string start
| - or
(?<=.)(?=\\p{Lu}) - any location between any char but linebreak chars and an uppercase letter.
Here is an altenative with a single call to gsubfn regex with some ifelse logic:
> gsubfn("^\\p{Ll}*(\\p{L})|(?<=.)(?=\\p{Lu})", function(n) ifelse(nchar(n)>0,n," "), x, perl=TRUE,backref=-1)
[1] "Text I Want To Display With Spacesmarch"
Here, the ^\\p{Ll}*(\\p{L}) part matches 0+ lowercase letters and captures the next uppercase into Group 1 that will be accessed by passing n argument to the anonymous function. If n length is non-zero, this alternative matched and the we need to replace with this value. Else, we replace with a space.
Since this is tagged perl, my 2 cents:
Can you chain together the substitutions inside sub() and gsub()? In newer perl versions an /r option can be added to the s/// substitution so the matched string can be returned "non-destructively" and then matched again. This allows hackish match/substitution/rematches without mastering advanced syntax, e.g.:
perl -E '
say "marchTextIWantToDisplayWithSpacesmarch" =~
s/\Amarch//r =~ s/([[:upper:]])/ $1/gr =~ s/\A\s//r;'
Output
Text I Want To Display With Spacesmarch
This seems to be what #pushpesh-kumar-rajwanshi and #akrun are doing by wrapping gsub inside sub() (and vice versa). In general I don't thinkperl = T captures the full magnificently advanced madness of perl regexps ;-) but gsub/sub must be fast operating on vectors, no?
The goal is to remove all non-capital letter in a string and I managed to find a regular expression solution without fully understanding it.
> gsub("[^::A-Z::]","", "PendingApproved")
[1] "PA"
I tried to read the documentation of regex in R but the double colon isn't really covered there.
[]includes characters to match in regex, A-Z means upper case and ^ means not, can someone help me understand what are the double colons there?
As far as I know, you don't need those double colons:
gsub("[^A-Z]", "", "PendingApproved")
[1] "PA"
Your current pattern says to remove any character which is not A-Z or colon :. The fact that you repeat the colons twice, on each side of the character range, does not add any extra logic.
Perhaps the author of the code you are using confounded the double colons with R's regex own syntax for named character classes. For example, we could have written the above as:
gsub("[^[:upper:]]","", "PendingApproved")
where [:upper:] means all upper case letters.
Demo
To remove all small letters use following:
gsub("[a-z]","", "PendingApproved")
^ denotes only starting characters so
gsub("^[a-z]","", "PendingApproved")
will not remove any letters from your tested string because your string don't have any small letters in starting of it.
EDIT: As per Tim's comment adding negation's work in character class too here. So let's say we want to remove all digits in a given value among alphabets and digits then following may help.
gsub("[^[:alpha:]]","", "PendingApproved1213133")
Where it is telling gsub then DO NOT substitute alphabets in this process. ^ works as negation in character class.
We can use str_remove from stringr
library(stringr)
str_remove_all("PendingApproved", "[a-z]+")
#[1] "PA"